RE: [sv-bc] Concat including enum as lvalue?

Jonathon,

Enums are treated the same as unpacked structures for matching and
equivalence rules. See 6.9.1 d). Section 6.9.3 defines assignment
compatibility for those types that define an implicit cast, and an enum
type defines an implicit cast in *one direction* only. See 4.10.3.

Therefore,

{B2, E2} = B4;

would require an implicit cast from a packed integral type to an enum to
be legal - no implicit cast is defined.

{B2, E2} = {2'b0, E2};

Is also illegal because the RHS creates an integral type 4-bits wide. It
is not an assignment pattern.

Dave


BTW, I noticed that 6.9.2b is superfluous. It is covered by 6.9.1c
together with 6.9.2a.


> -----Original Message-----
> From: owner-sv-bc@server.eda-stds.org [mailto:owner-sv-bc@server.eda-
> stds.org] On Behalf Of Jonathan Bromley
> Sent: Tuesday, June 20, 2006 6:57 AM
> To: sv-bc@server.eda-stds.org
> Subject: [sv-bc] Concat including enum as lvalue?
> 
> Once again I beg a clarification, with the usual caveat that
> the answer may already be in the LRM and I've just missed it.
> If that's the case, feel free to slap my wrists.
> 
> Enum variables are strongly typed in the sense that if
> used as the target of an assignment, the expression
> forming the right-hand side of the assignment must be
> of the same enum type.  So far, so good.
> 
> Whenever an enum appears as an expression, its numeric
> value may be used without explicit casting.  Also OK,
> if perhaps a little free-and-easy.
> 
> So, therefore, I think the following is OK:
> 
>   enum bit[1:0] {A, B, C, D} E2;
>   bit [1:0] B2;
>   bit [3:0] B4;
>   ...
>   B4 = {B2, E2};
> 
> because E2 is silently cast to "bit [1:0]" for
> use in the expression.  But what about
> 
>   {B2, E2} = B4;
> 
> ??  The tools I've tried permit this, although at least
> one of them can be persuaded to generate mysteriously
> wrong answers.  I can see reasonable arguments both for
> and against accepting such a construct, but I don't
> see the LRM pronouncing on it.
> 
> If there *is* strong type checking in such a case, then
> is the following OK?
> 
>   {B2, E2} = {2'b0, E2};  // component types match
> 
> Thanks
> --
> Jonathan Bromley, Consultant
> 
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